1984 AIME Problems/Problem 11

Revision as of 18:29, 19 August 2019 by Nafer (talk | contribs) (Solution 2)


A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$.


First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and "non-birch" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as opposed to those with few branches. Indeed, you could keep dividing until you have them each in their own category, but in the end it will not change the probability of the birch trees being near each other. That is, in the end, you multiply the numerator by the number of ways to arrange the oak and maple trees and you also multiply the denominator by the number of ways to arrange the oak and maple trees, making them cancel out.)

The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 5 dividers of 8 slots that the birch trees can go in, making ${8\choose5} = 56$ different ways to arrange this.

There are ${12 \choose 5} = 792$ total ways to arrange the twelve trees, so the probability is $\frac{56}{792} = \frac{7}{99}$.

The answer is $7 + 99 = \boxed{106}$.

Solution 2

Let $b$, $n$ denote birch tree and not-birch tree, respectively. Notice that we only need $4$ $n$s to separate the $5$ $b$s. Specifically, \[b,n,b,n,b,n,b,n,b\] Since we have $7$ $n$s, we are placing the extra $3$ $n$s into the intervals \[_bbbbb\]

See also

1984 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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